# Find the number of group homomorphism between $A_5$ and $S_5$.

The above question is based on this answer to a similar question. I just want to apply what has been pointed out in that answer to this question.

So we are interested in number of homomorphisms $$f:A_5 \to S_5$$ (if any). Now following the ideas given in the linked answer, $$A_5$$ is simple, so it has to normal subgroups, viz $$\{e\}$$ and $$A_5$$.

If $$f$$ is a homomorphism then ker$$f$$ is a normal subgroup of $$A_5$$.

1. If ker$$f$$=$$\{e\}$$ , then $$A_5/\{e\}\sim f(A_5)$$, and so $$f(A_5)$$ is a subgroup of $$S_5$$ of order $$60$$.
2. If ker$$f=A_5$$, then $$f$$ is trivial.

Also we know that "Any homomorphism from a simple group to any group is either trivial or injective."

Now in the view of above argument what can I conclude? It seems to me that only I can conclude that any homomorphism is injective. But how to calculate how many are there? Is it a valid approach or do we need a new approach?

Thanks.

• Also we know that "Any homomorphism from a simple group to any group is either trivial or injective." $-$ You have shown that already in $(1)$ and $(2).$ Mar 4 '19 at 7:38
• I think $A_5$ is the only subgroup of $S_5$ of order $60.$ So in order to find the number of injective homomorphisms from $A_5$ to $S_5$ you need only to find the number of elements in $\text {Aut} (A_5) = \left |A_5/Z(A_5) \right| = |A_5| = 60,$ since $Z(A_5) = \{e \},$ where $Z(A_5)$ is the center of $A_5.$ Mar 4 '19 at 7:48
• For the proof of the above fact see here math.stackexchange.com/q/822096/543867 Mar 4 '19 at 7:59
• "If ker$f$=$\{e\}$ , then $A_5/\{e\}\sim f(A_5)$" What does "$\sim$" mean in this context? Mar 4 '19 at 8:15

As you observed, the non-constant homomorphisms are injective. Because $$A_5$$ is the only subgroup of $$S_5$$ of order $$60$$ we are looking for automorphisms from $$A_5$$ to itself.

To each element $$g\in S_5$$ we get a conjugation automorphism $$\phi_g(x)=gxg^{-1}$$ for all $$x\in A_5$$. Because the centralizer of $$A_5$$ in $$S_5$$ is trivial, distinct choices of $$g$$ yield distinct automorphisms $$\phi_g$$.

Claim. Any automorphism $$\phi$$ of $$A_5$$ is of the form $$\phi_g$$ for some $$g\in S_5$$.

Proof. The group $$A_5$$ has five distinct Sylow $$2$$-subgroups. Namely $$P_5=\{1,(12)(34),(13)(24),(14)(23)\}$$ and its conjugates, each stabilizing a single element of $$J_5:=\{1,2,3,4,5\}$$. I will denote by $$P_i$$ the conjugate stabilizing $$i\in J_5$$. Because $$\phi$$ is an automorphism it must permute these 5 groups. So there is a permutation $$\sigma\in S_5$$ such that $$\phi(P_i)=P_{\sigma(i)}$$ for all $$i\in J_5$$.

On the other hand, the conjugation $$\phi_\sigma$$ also maps $$P_i$$ to $$P_{\sigma(i)}$$. Therefore the automorphism $$\tau:=\phi\circ\phi_{\sigma^{-1}}$$ has the property that $$\tau(P_i)=P_i$$ for all $$i\in J_5$$. Consequently also the normalizers are preserved: $$\tau(N_{A_5}(P_i)=N_{A_5}(P_i)$$ for all $$i\in J_5$$. But those normalizers are conjugates of $$A_4$$, each the stabilizer of an element of $$J_5$$. Consider a 3-cycle, such as $$\alpha=(234)$$. The only two 3-cycles normalizing both $$P_1$$ and $$P_5$$ are $$\alpha$$ and $$\alpha^{-1}=(243)$$. Therefore we must have $$\tau(\alpha)=\alpha$$ or $$\tau(\alpha)=\alpha^{-1}$$. But, we have $$\tau(\alpha P_2\alpha^{-1})=\tau(P_3)=P_3$$ as well as $$\tau(\alpha P_2\alpha^{-1})=\tau(\alpha)P_2\tau(\alpha)^{-1}=P_{\tau(\alpha)(2)}.$$ So we must have $$\tau(\alpha)(2)=3$$, leaving $$\tau(\alpha)=\alpha$$ as the only possibility.

It follows that $$\tau(\beta)=\beta$$ for all 3-cycles $$\beta\in A_5$$. But the 3-cycles generate $$A_5$$, so $$\tau$$ must be the identity mapping. Therefore $$\phi=\phi_{\sigma}$$. QED.

It follows that there are 120 injective homomorphisms from $$A_5$$ to $$S_5$$ and the trivial constant homomorphism.

• What was going wrong in my argument in the comments above @Jyrki Lahtonen? Mar 4 '19 at 8:09
• The same argument goes thru for $A_n$, $n\ge7$. The case $n=6$ is a famous exception for $S_6$ when the inner automorphisms form an index two subgroup of the automorphism group. See here for $n\ge7$. Mar 4 '19 at 8:09
• Oh! Sorry. I have calculated order of $\text {Inn} (G).$ Right? Mar 4 '19 at 8:12
• @Dbchatto67 I am not aware of one. I suspect it is difficult in general. Let's wait for an expert. Mar 4 '19 at 8:16
• Possibly @KushalBhuyan. If you can prove some other way that the obvious conjugates of $A_4$ are the only subgroups isomorphic to $A_4$, and hence $\phi$ must permute them. The reason why I used Sylow 2-subgroups is that, as a set, they are characteristic to the group $A_5$, and hence any automorphism must permute them. Mar 4 '19 at 11:14